Dr. Hackney STA Solutions pg 157

Dr. Hackney STA Solutions pg 157 - Second Edition 9-15...

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Unformatted text preview: Second Edition 9-15 where c = 2 / 2 . The derivative (with respect to c) of this last expression is bfn-1 (bc) - afn-1 (ac), and hence is equal to zero if both c = 1 (so the interval is unbiased) and bfn-1 (b) = afn-1 (a). From the form of the chi squared pdf, this latter condition is equivalent to fn+1 (b) = fn+1 (a). d. By construction, the interval will be 1 - equal-tailed. 9.53 a. E [blength(C) - IC ()] = 2cb - P (|Z| c), where Z n(0, 1). b. [2cb - P (|Z| c)] = 2b - 2 1 e-c /2 . 2 2 c. If b > 1/ 2 the derivative is always positive since e-c /2 < 1. d dc 2 9.55 E[L((,), C)] = E [L((,), C)|S < K] P (S < K) + E [L((,), C)|S > K] P (S > K) = E L((,), C )|S < K P (S < K) + E [L((,), C)|S > K] P (S > K) = R L((,), C ) + E [L((,), C)|S > K] P (S > K), where the last equality follows because C = if S > K. The conditional expectation in the second term is bounded by E [L((,), C)|S > K] = = > = E [blength(C) - IC ()|S > K] E [2bcS - IC ()|S > K] E [2bcK - 1|S > K] (since S > K and IC 1) 2bcK - 1, which is positive if K > 1/2bc. For those values of K, C dominates C. 9.57 a. The distribution of Xn+1 - X is n[0, 2 (1 + 1/n)], so P Xn+1 X z/2 1 + 1/n = P (|Z| z/2 ) = 1 - . b. p percent of the normal population is in the interval zp/2 , so x k is a 1 - tolerance interval if P ( zp/2 X k) = P (X - k - zp/2 and X + k + zp/2 ) 1 - . This can be attained by requiring P (X - k - zp/2 ) = /2 and P (X + k + zp/2 ) = /2, which is attained for k = zp/2 + z/2 / n. c. From part (a), (Xn+1 - X)/(S 1 + 1/n) tn-1 , so a 1 - prediction interval is X tn-1,/2 S 1 + 1/n. ...
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